100-seat airplane problem
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12 members have voted
March 5th, 2020 at 6:12:58 AM
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I've been asked to ask a math question was wasn't a geometry problem. Here you go. This one is an oldie but goody. I may have asked it before.
An airplane has 100 seats and there are 100 passengers in a line, waiting to board. Every passenger has an assigned seat. The first passenger will ignore his assigned seat and pick one at random. The other 99 passengers have the following behavior:
1. He will look for his seat and sit in it if it's empty.
2. If the assigned seat is occupied, he will pick a random empty seat.
What is the probability the last passenger to board sits in his assigned seat?
Usual rules apply, please.
An airplane has 100 seats and there are 100 passengers in a line, waiting to board. Every passenger has an assigned seat. The first passenger will ignore his assigned seat and pick one at random. The other 99 passengers have the following behavior:
1. He will look for his seat and sit in it if it's empty.
2. If the assigned seat is occupied, he will pick a random empty seat.
What is the probability the last passenger to board sits in his assigned seat?
Usual rules apply, please.
Last edited by: Wizard on Mar 5, 2020
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
March 5th, 2020 at 6:33:37 AM
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I can't solve it. LOL
Edited
May be can use simulation to do it.
Edited
May be can use simulation to do it.
March 5th, 2020 at 7:13:31 AM
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I’ll just say, highly unlikely.
A falling knife has no handle.
March 5th, 2020 at 7:58:38 AM
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Easy.
The seat is either occupied or not, makes it 50%.
:-)
The seat is either occupied or not, makes it 50%.
:-)
March 5th, 2020 at 9:37:36 AM
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Do the pilots stand? If the plane only has 100 seats and there are 100 passengers, someone will be standing.
At my age, a "Life In Prison" sentence is not much of a deterrent.
March 5th, 2020 at 10:08:13 AM
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I love this problem. Will refrain from answering since I know it.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
March 5th, 2020 at 10:40:48 AM
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Quote: DRichDo the pilots stand? If the plane only has 100 seats and there are 100 passengers, someone will be standing.
Yes, the pilot has to fly the plane standing up.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
March 5th, 2020 at 10:49:21 AM
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This was a thread a few years ago, wasn't it?
March 5th, 2020 at 11:44:07 AM
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I looked this up. I would appreciate some help understanding the solution, after it has been answered. I’m in the middle of a foggy mind day.
A falling knife has no handle.
March 5th, 2020 at 12:10:48 PM
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Quote: rsactuaryThis was a thread a few years ago, wasn't it?
Probably. However, I'm sure nobody everybody had the pleasure of solving it.
Quote: MoscaI looked this up. I would appreciate some help understanding the solution, after it has been answered. I’m in the middle of a foggy mind day.
To be honest, I'm not sure my solution would hold up under strict mathematical scrutiny, but does pass my common sense test.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
March 5th, 2020 at 1:14:09 PM
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Interesting.
I wrote a quick & dirty simulation in Excel to simulate this problem and I'm getting an answer of 50%.
First I just tried the problem with 10 seats, not 100. (since I could run the sim many more times... it wouldn't take so long to run) and that gave an answer near 50% after a million or so sims. I then modified the code to simulate the problem with the full 100 seats and yet that also gives an answer of 50%.
Maybe something's wrong with my code but I've coded hundred and hundreds of simulations. So far I don't see anything wrong with my source code.
March 5th, 2020 at 1:58:07 PM
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Quote: EdCollins
Interesting.
I wrote a quick & dirty simulation in Excel to simulate this problem and I'm getting an answer of 50%.
First I just tried the problem with 10 seats, not 100. (since I could run the sim many more times... it wouldn't take so long to run) and that gave an answer near 50% after a million or so sims. I then modified the code to simulate the problem with the full 100 seats and yet that also gives an answer of 50%.
Maybe something's wrong with my code but I've coded hundred and hundreds of simulations. So far I don't see anything wrong with my source code.
I agree the answer is 50%. I always find simulations so unsatisfying from an intellectual point of view. For the beer, so I'd like to see something more mathematical.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
March 5th, 2020 at 2:14:48 PM
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Quote: Wizard
I agree the answer is 50%. I always find simulations so unsatisfying from an intellectual point of view. For the beer, so I'd like to see something more mathematical.
I agree, but sadly my coding skills far exceed my mathematical abilities.
At my age, a "Life In Prison" sentence is not much of a deterrent.
March 5th, 2020 at 2:17:20 PM
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Me too!!!!Quote: DRichQuote: Wizard
I agree the answer is 50%. I always find simulations so unsatisfying from an intellectual point of view. For the beer, so I'd like to see something more mathematical.
I agree, but sadly my coding skills far exceed my mathematical abilities.
March 5th, 2020 at 2:53:14 PM
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Nice puzzle - PM sent.Quote: Wizard...What is the probability the last passenger to board sits in his assigned seat?...
March 5th, 2020 at 3:42:18 PM
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Quote: DRichQuote: Wizard
I agree the answer is 50%. I always find simulations so unsatisfying from an intellectual point of view. For the beer, so I'd like to see something more mathematical.
I agree, but sadly my coding skills far exceed my mathematical abilities.
There’s a pretty simple logical explanation for that answer. But I’ll refrain from posting it for now. I will say that it was quite cool to realize it the first time I worked out the solution.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
March 5th, 2020 at 4:20:45 PM
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My simulation results, 1 million trials, probability = 0.50002
Actually, the probability will remain equal to 0.5 for ANY no of passengers(>=2) !
I will post the vba code if anyone interested.
https://imge.to/i/yNjRf2
I totally agree that mathematical solutions is much better than simulation.
Anyone here can teach me the mathematical solutions of this problem ?
Actually, the probability will remain equal to 0.5 for ANY no of passengers(>=2) !
I will post the vba code if anyone interested.
https://imge.to/i/yNjRf2
I totally agree that mathematical solutions is much better than simulation.
Anyone here can teach me the mathematical solutions of this problem ?
Last edited by: ssho88 on Mar 5, 2020
March 5th, 2020 at 7:08:48 PM
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Okay, I think I got it figured out why. Each time a passenger makes a choice, it is either the 100th passenger’s seat, or one of a much larger array. As long as that 100th seat isn’t taken earlier, then it is 1/2. It might be taken earlier, but the chance of it being taken 1st is 1/100, 2nd is 1/99, 3rd is 1/98, etc.
Or, do it backwards... start with 2 seats; 1/2. If the first passenger chooses the off-seat, add a chair; still 1/2. Continue until you hit 100.
Or, do it backwards... start with 2 seats; 1/2. If the first passenger chooses the off-seat, add a chair; still 1/2. Continue until you hit 100.
A falling knife has no handle.
March 5th, 2020 at 7:29:00 PM
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Here is an argument why it is 0.5.
The final remaining seat must be the one that was assigned to either passenger 1 or passenger 100. No other assigned seat can be empty at the end, because the matching passenger would have taken it when they got on the plane. So only those two seats are possible.
Every time a passenger picks a random seat, they are equally likely to pick seat 1 or seat 100. So those two seats are equally likely to be unselected all the way through.
The final remaining seat must be the one that was assigned to either passenger 1 or passenger 100. No other assigned seat can be empty at the end, because the matching passenger would have taken it when they got on the plane. So only those two seats are possible.
Every time a passenger picks a random seat, they are equally likely to pick seat 1 or seat 100. So those two seats are equally likely to be unselected all the way through.
March 5th, 2020 at 7:39:36 PM
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Quote: MrGoldenSunHere is an argument why it is 0.5.
The final remaining seat must be the one that was assigned to either passenger 1 or passenger 100. No other assigned seat can be empty at the end, because the matching passenger would have taken it when they got on the plane. So only those two seats are possible.
That's good. I never realized the last seat must be #1 or #100. I'm not sure I could convince somebody else the odds are 50/50. Let me chew on this.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
March 5th, 2020 at 7:53:28 PM
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This is why I fly Southwest
March 5th, 2020 at 8:18:51 PM
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The first passenger can seal the fate if he takes seat 1 or seat 100. Those are equally likely. Then if he picked anything else, it's a coin flip from now on as to which gets taken first. And once one is taken, the fate is sealed.
March 5th, 2020 at 8:23:14 PM
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Quote: MrGoldenSunThe first passenger can seal the fate if he takes seat 1 or seat 100. Those are equally likely. Then if he picked anything else, it's a coin flip from now on as to which gets taken first. And once one is taken, the fate is sealed.
That's correct ! Thanks for your explanation ! The possible seat for last passenger is either seat #1 or seat #100 !
There are 3 scenarios here :-
1) if Passenger 1 pick seat #1, then Passenger 100 will definitely pick seat #100
2) if Passenger 1 pick seat #100, then Passenger 100 will definitely pick seat #1
3) if Passenger 1 pick any other seat #n, then one of the Passenger 2 to Passenger 99 will eventually pick either seat #1 or seat #100.
If that passenger pick seat #1, again passenger 100 will definitely pick seat #100, if that passenger pick seat #100, passenger 100 will definitely pick seat #1.
So the possible seat for last passenger is either seat #1 or seat #100 !
Last edited by: ssho88 on Mar 5, 2020
March 5th, 2020 at 10:12:09 PM
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I believe this is the same math problem for which I won the beer a couple of years ago.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
March 5th, 2020 at 11:15:18 PM
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Nice puzzle, I haven't seen it before so had come up with this approach which seems to explain the 50% idea and mathematically sound.
Part 1)
Initially #1 could (a) pick seat 1 or (b) seat 100 with equal probability. Seat 1 leads to everyone picking their own seats, so works. Seat 100 means it's impossible for #100 to be right. (c) Anything else #1 sits in seat M. Go to Part 2).
Part 2)
People from 2 to M-1 all pick their own seats, so now consider when #M has to choose a seat. Seat 1 is empty (since #1 is sitting in seat M), all the seats 2 to M-1 are full, all the seats M+1 to 100 are empty. There are three possibilities.
(a) #M sits in seat 1, thereafter everyone will pick their own seats, so works.
(b) #M sits in seat 100, so now impossible for #100 to be right.
(c) #M sits in seat N (M<N<100). This is recursion so repeat the process, i.e. Part 2), using N.
Part 3)
Under Part 2, if there is no outcome, N keeps increasing. Eventually there's either an outcome or N reaches 99. If N reaches 99 then there are only two seats free, 1 or 100. So either #99 picks (a) seat 1 or (b) seat 100, thus a 50:50 chance of success.
Summary
At every stage where there's an outcome it's 50:50 whether you get (a) success or (b) failure, otherwise N keeps increasing. If it gets to 99, as shown in part 3), there's an outcome. So whatever happens there will eventually be an outcome and each outcome is 50:50.
Thus the chance of success is 50%.
Part 1)
Initially #1 could (a) pick seat 1 or (b) seat 100 with equal probability. Seat 1 leads to everyone picking their own seats, so works. Seat 100 means it's impossible for #100 to be right. (c) Anything else #1 sits in seat M. Go to Part 2).
Part 2)
People from 2 to M-1 all pick their own seats, so now consider when #M has to choose a seat. Seat 1 is empty (since #1 is sitting in seat M), all the seats 2 to M-1 are full, all the seats M+1 to 100 are empty. There are three possibilities.
(a) #M sits in seat 1, thereafter everyone will pick their own seats, so works.
(b) #M sits in seat 100, so now impossible for #100 to be right.
(c) #M sits in seat N (M<N<100). This is recursion so repeat the process, i.e. Part 2), using N.
Part 3)
Under Part 2, if there is no outcome, N keeps increasing. Eventually there's either an outcome or N reaches 99. If N reaches 99 then there are only two seats free, 1 or 100. So either #99 picks (a) seat 1 or (b) seat 100, thus a 50:50 chance of success.
Summary
At every stage where there's an outcome it's 50:50 whether you get (a) success or (b) failure, otherwise N keeps increasing. If it gets to 99, as shown in part 3), there's an outcome. So whatever happens there will eventually be an outcome and each outcome is 50:50.
Thus the chance of success is 50%.
March 6th, 2020 at 6:17:39 AM
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Good replies!
When I first did this years ago, I noticed the answer was 50% for 2 to 5 or so. I figured that was enough to show a pattern and felt satisfied the answer for any number of passengers greater than 1 was 50%.
When I first did this years ago, I noticed the answer was 50% for 2 to 5 or so. I figured that was enough to show a pattern and felt satisfied the answer for any number of passengers greater than 1 was 50%.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
March 6th, 2020 at 6:20:07 AM
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Quote: charliepatrickNice puzzle, I haven't seen it before so had come up with this approach which seems to explain the 50% idea and mathematically sound.
Part 1)
Initially #1 could (a) pick seat 1 or (b) seat 100 with equal probability. Seat 1 leads to everyone picking their own seats, so works. Seat 100 means it's impossible for #100 to be right. (c) Anything else #1 sits in seat M. Go to Part 2).
Part 2)
People from 2 to M-1 all pick their own seats, so now consider when #M has to choose a seat. Seat 1 is empty (since #1 is sitting in seat M), all the seats 2 to M-1 are full, all the seats M+1 to 100 are empty. There are three possibilities.
(a) #M sits in seat 1, thereafter everyone will pick their own seats, so works.
(b) #M sits in seat 100, so now impossible for #100 to be right.
(c) #M sits in seat N (M<N<100). This is recursion so repeat the process, i.e. Part 2), using N.
Part 3)
Under Part 2, if there is no outcome, N keeps increasing. Eventually there's either an outcome or N reaches 99. If N reaches 99 then there are only two seats free, 1 or 100. So either #99 picks (a) seat 1 or (b) seat 100, thus a 50:50 chance of success.
Summary
At every stage where there's an outcome it's 50:50 whether you get (a) success or (b) failure, otherwise N keeps increasing. If it gets to 99, as shown in part 3), there's an outcome. So whatever happens there will eventually be an outcome and each outcome is 50:50.
Thus the chance of success is 50%.
Yes exactly agree. The game either effectively ends in round 1 if seat 1 or 100 are randomly picked (and if it ends, it’s 50/50 winner or loser). Otherwise it becomes an identical game of 100 - X seats.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
March 6th, 2020 at 11:36:44 AM
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Interesting.
Based on the prior replies, I agree that it's 50%.
It's a closed loop. There may be any number of displaced passengers, but once somebody sits in seat #1 or seat #100, it's game over. 50%
Bottom line, it's a logic problem, not a math problem.
But it kinda presents a new problem. And I think this IS a math problem:
On average, how many passengers are in the wrong seat?
My knee-jerk answer, based on nothing really, is 50.
Based on the prior replies, I agree that it's 50%.
It's a closed loop. There may be any number of displaced passengers, but once somebody sits in seat #1 or seat #100, it's game over. 50%
Bottom line, it's a logic problem, not a math problem.
But it kinda presents a new problem. And I think this IS a math problem:
On average, how many passengers are in the wrong seat?
My knee-jerk answer, based on nothing really, is 50.
I invented a few casino games. Info:
http://www.DaveMillerGaming.com/ —————————————————————————————————————
Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
March 6th, 2020 at 1:30:58 PM
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I misunderstood the problem. I did mention that I was having a foggy mind day.
A falling knife has no handle.
March 6th, 2020 at 2:53:18 PM
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Quote: DJTeddyBearBut it kinda presents a new problem. And I think this IS a math problem:
On average, how many passengers are in the wrong seat?
Just 5.18 I think.
March 6th, 2020 at 8:55:55 PM
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Quote: MrGoldenSunHere is an argument why it is 0.5.
The final remaining seat must be the one that was assigned to either passenger 1 or passenger 100. No other assigned seat can be empty at the end, because the matching passenger would have taken it when they got on the plane. So only those two seats are possible.
Every time a passenger picks a random seat, they are equally likely to pick seat 1 or seat 100. So those two seats are equally likely to be unselected all the way through.
I don't think this is right and proves the result is 0.5
The empty seat at the end can be any seat. Example, passenger 1 picks seat 10. passengers 2 to 9 get their seat. Passenger 10 randomly picks seat 100. Passengers 11 to 99 get their seat. Passenger 100 gets seat 10.
Quote: charliepatrickNice puzzle, I haven't seen it before so had come up with this approach which seems to explain the 50% idea and mathematically sound.
Part 1)
Initially #1 could (a) pick seat 1 or (b) seat 100 with equal probability. Seat 1 leads to everyone picking their own seats, so works. Seat 100 means it's impossible for #100 to be right. (c) Anything else #1 sits in seat M. Go to Part 2).
Part 2)
People from 2 to M-1 all pick their own seats, so now consider when #M has to choose a seat. Seat 1 is empty (since #1 is sitting in seat M), all the seats 2 to M-1 are full, all the seats M+1 to 100 are empty. There are three possibilities.
(a) #M sits in seat 1, thereafter everyone will pick their own seats, so works.
(b) #M sits in seat 100, so now impossible for #100 to be right.
(c) #M sits in seat N (M<N<100). This is recursion so repeat the process, i.e. Part 2), using N.
Part 3)
Under Part 2, if there is no outcome, N keeps increasing. Eventually there's either an outcome or N reaches 99. If N reaches 99 then there are only two seats free, 1 or 100. So either #99 picks (a) seat 1 or (b) seat 100, thus a 50:50 chance of success.
Summary
At every stage where there's an outcome it's 50:50 whether you get (a) success or (b) failure, otherwise N keeps increasing. If it gets to 99, as shown in part 3), there's an outcome. So whatever happens there will eventually be an outcome and each outcome is 50:50.
Thus the chance of success is 50%.
This is rigorous! Not that you need me to tell you that ^_^"
March 6th, 2020 at 10:19:19 PM
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Quote: tyler498I don't think this is right and proves the result is 0.5
The empty seat at the end can be any seat. Example, passenger 1 picks seat 10. passengers 2 to 9 get their seat. Passenger 10 randomly picks seat 100. Passengers 11 to 99 get their seat. Passenger 100 gets seat 10.Quote: charliepatrickNice puzzle, I haven't seen it before so had come up with this approach which seems to explain the 50% idea and mathematically sound.
Part 1)
Initially #1 could (a) pick seat 1 or (b) seat 100 with equal probability. Seat 1 leads to everyone picking their own seats, so works. Seat 100 means it's impossible for #100 to be right. (c) Anything else #1 sits in seat M. Go to Part 2).
Part 2)
People from 2 to M-1 all pick their own seats, so now consider when #M has to choose a seat. Seat 1 is empty (since #1 is sitting in seat M), all the seats 2 to M-1 are full, all the seats M+1 to 100 are empty. There are three possibilities.
(a) #M sits in seat 1, thereafter everyone will pick their own seats, so works.
(b) #M sits in seat 100, so now impossible for #100 to be right.
(c) #M sits in seat N (M<N<100). This is recursion so repeat the process, i.e. Part 2), using N.
Part 3)
Under Part 2, if there is no outcome, N keeps increasing. Eventually there's either an outcome or N reaches 99. If N reaches 99 then there are only two seats free, 1 or 100. So either #99 picks (a) seat 1 or (b) seat 100, thus a 50:50 chance of success.
Summary
At every stage where there's an outcome it's 50:50 whether you get (a) success or (b) failure, otherwise N keeps increasing. If it gets to 99, as shown in part 3), there's an outcome. So whatever happens there will eventually be an outcome and each outcome is 50:50.
Thus the chance of success is 50%.
This is rigorous! Not that you need me to tell you that ^_^"
If Passenger 10 randomly picks seat 100, then Passenger 100 will definitely picks seat 1, ^_^"
March 7th, 2020 at 7:19:13 AM
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Quote: ssho88
I will post the vba code if anyone interested.
Yes please. It's more about using and getting familiar with vba than the actual question at hand.
March 7th, 2020 at 8:31:17 AM
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Quote: DJTeddyBearBut it kinda presents a new problem. And I think this IS a math problem:
On average, how many passengers are in the wrong seat?
My knee-jerk answer, based on nothing really, is 50.
1/100 of the time, passenger 1 is in seat 100, so there are only 2 in the wrong seats (1 and 100 in each other's seat)
1/100 of the time, passenger 1 is in seat 99
In this case, 1/2 of the time, passenger 99 is in seat 1 (2 wrong), and 1/2 of the time, 99 is in 100 and 100 is in 1 (3 wrong), which is a mean of 5/2
1/100 of the time, passenger 1 is in seat 98
In this case, 1/3 of the time, passenger 98 is in seat 1 (2 wrong); 1/3 of the time, 98 is in 100 and 100 is in 1 (3 wrong); and 1/3 of the time, 98 is in 99, resulting in an additional expected 5/2 wrong; this is a total of 1/3 (2 + 3 + 5/2) = 17/6
Continue doing this until all 100 seats for passenger 1 has been considered; the answer is:
1/100 (0 + 2 + (2 + 1/2) + (2 + 1/2 + 1/3) + (2 + 1/2 + 1/3 + 1/4) + ... + (2 + 1/2 + 1/3 + ... + 1/99)) = about 5.1773775
March 7th, 2020 at 9:11:10 AM
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Quote: ssho88Quote: tyler498I don't think this is right and proves the result is 0.5
The empty seat at the end can be any seat. Example, passenger 1 picks seat 10. passengers 2 to 9 get their seat. Passenger 10 randomly picks seat 100. Passengers 11 to 99 get their seat. Passenger 100 gets seat 10.Quote: charliepatrickNice puzzle, I haven't seen it before so had come up with this approach which seems to explain the 50% idea and mathematically sound.
Part 1)
Initially #1 could (a) pick seat 1 or (b) seat 100 with equal probability. Seat 1 leads to everyone picking their own seats, so works. Seat 100 means it's impossible for #100 to be right. (c) Anything else #1 sits in seat M. Go to Part 2).
Part 2)
People from 2 to M-1 all pick their own seats, so now consider when #M has to choose a seat. Seat 1 is empty (since #1 is sitting in seat M), all the seats 2 to M-1 are full, all the seats M+1 to 100 are empty. There are three possibilities.
(a) #M sits in seat 1, thereafter everyone will pick their own seats, so works.
(b) #M sits in seat 100, so now impossible for #100 to be right.
(c) #M sits in seat N (M<N<100). This is recursion so repeat the process, i.e. Part 2), using N.
Part 3)
Under Part 2, if there is no outcome, N keeps increasing. Eventually there's either an outcome or N reaches 99. If N reaches 99 then there are only two seats free, 1 or 100. So either #99 picks (a) seat 1 or (b) seat 100, thus a 50:50 chance of success.
Summary
At every stage where there's an outcome it's 50:50 whether you get (a) success or (b) failure, otherwise N keeps increasing. If it gets to 99, as shown in part 3), there's an outcome. So whatever happens there will eventually be an outcome and each outcome is 50:50.
Thus the chance of success is 50%.
This is rigorous! Not that you need me to tell you that ^_^"
If Passenger 10 randomly picks seat 100, then Passenger 100 will definitely picks seat 1, ^_^"
oh right, my bad ^_^" brain freeze..
